The lengths of the sides of a triangle are 13 | vedclass

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(B) Let the sides of the triangle be $a = 13 \, cm$,$b = 13 \, cm$,and $c = 10 \, cm$.First,calculate the semi-perimeter $s = (a + b + c) / 2 = (13 +

Vedclass · Accepted Answer

$60 \, cm^2$,$12 \, cm$

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(B) Let the sides of the triangle be $a = 13 \, cm$,$b = 13 \, cm$,and $c = 10 \, cm$.First,calculate the semi-perimeter $s = (a + b + c) / 2 = (13 + 13 + 10) / 2 = 36 / 2 = 18 \, cm$.Using Heron's formula,Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{18(18-13)(18-13)(18-10)} = \sqrt{18 	imes 5 	imes 5 	imes 8} = \sqrt{3600} = 60 \, cm^2$.Now,to find the altitude $h$ corresponding to the base $c = 10 \, cm$,use the formula Area $= (1/2) 	imes 	ext{base} 	imes 	ext{height}$.$60 = (1/2) 	imes 10 	imes h \implies 60 = 5h \implies h = 12 \, cm$.

The lengths of the sides of a triangle are $13 \, cm$,$13 \, cm$,and $10 \, cm$. Find its area. Also,find the length of the altitude from the opposite vertex to the side whose length is $10 \, cm$.

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